The family of circles x2 + y2 + ax + ay = 2 + 2a for different values of 'a' pass
(a) One fixed point
(b) Two fixed point
(c) More than two fixed point
(d) None of these
Answers
Answer:
non of these
Step-by-step explanation
The family of circles x² + y² + ax + ay = 2 + 2a for different values of 'a' pass will pass through two fixed points (-√2 , - √2) & (√2 , √2)
Step-by-step explanation:
x² + y² + ax + ay = 2 + 2a
=> (x + a/2)² - a²/4 + (y + a/2)² - a²/4 = 2 + 2a
=> (x + a/2)² + (y + a/2)² = a²/4 + 2a + 2
=> (x + a/2)² + (y + a/2)² = (a/√2 + 2)²
Center = (-a/2 , -a/2) Radius = (a/√2 + 2)
if a = 0
then Circle center at origin will radius = 2
hence circle has point (-√2 , - √2) , (-√2 , √2) , (√2 , -√2) & (√2 , √2)
For a = + ve => cenetr lies in 3rd Quadrant
Distance of (√2 , √2) from center = √((√2 + a/2)² + (√2 + a/2)²)
= (2 + a√2) = Radius
For a = -ve => cenetr lies in 1st Quadrant
Distance of (-√2 , -√2) from center = √((-√2 - a/2)² + (√2 - a/2)²)
= (2 + a√2) = Radius
Hence The family of circles x² + y² + ax + ay = 2 + 2a for different values of 'a' pass will pass through two fixed points (-√2 , - √2) & (√2 , √2)
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